Divide the following complex numbers. $ \dfrac{5-10i}{4-3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4+3i}$ $ \dfrac{5-10i}{4-3i} = \dfrac{5-10i}{4-3i} \cdot \dfrac{{4+3i}}{{4+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(5-10i) \cdot (4+3i)} {(4-3i) \cdot (4+3i)} = \dfrac{(5-10i) \cdot (4+3i)} {4^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(5-10i) \cdot (4+3i)} {(4)^2 - (-3i)^2} = $ $ \dfrac{(5-10i) \cdot (4+3i)} {16 + 9} = $ $ \dfrac{(5-10i) \cdot (4+3i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({5-10i}) \cdot ({4+3i})} {25} = $ $ \dfrac{{5} \cdot {4} + {-10} \cdot {4 i} + {5} \cdot {3 i} + {-10} \cdot {3 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{20 - 40i + 15i - 30 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{20 - 40i + 15i + 30} {25} = \dfrac{50 - 25i} {25} = 2-i $